Torque in Gears 
An input force $F$ is applied to gear A which causes a torque $T$.

*

*Is the torque on $B$ the same as $A$ i.e. equal to $T$ or

*Is the tangent force on $B$ same as the tangent force on $A$ i.e. equal to $F$
$F$ is being constantly supplied. Ignore friction.
 A: In general, neither statement $1$ nor statement $2$ is true (even if you assume friction is negligible). To determine the torque exerted on $B$ you will need to know the ratio of the number of teeth on $A$ to the number of teeth on $B$ and the ratio of their moments of inertia.
A: So neither of those statements are true in general. Here are the conditions that make each one true.
In the diagram there is a equal and opposite contact force $N$ between the two gears at the contact point. This force can be tangential to the gear, but in real life it is always pointing inwards. Also note the perpendicular distance (moment arm) of this force to each gear center.

The torque on A due to the contact is $\tau_A = N \ r_A$, and the torque on B due to the contact is $\tau_B = N \ r_B$.
$$ N = \frac{\tau_A}{r_A} = \frac{\tau_B}{r_B} $$
The only way this torque matches both sides is if the gears have the same moment arms $r_A=r_B$.
Now it is not necessary for the input torque $T$ to match with torque $\tau_A$. Why? Because if the gear is accelerating then some of that torque is lost
$$ T = I_A \alpha_A + \tau_A $$ is the equation of motion of Gear A.
You can translate the above in terms of the tangent force $F_A = T \ r_A$ and the contact force $N = \tau_A / r_A$
$$ F_A = \frac{I_A \alpha_A}{r_A} + N $$
Again the tangent force is only equal to the contact force if the gears are not accelerating.
Using Newton's 3rd we can see the contact force $N$ is also applied to Gear B.
A: If there is no friction, the torque in the gear A leads to angular acceleration.
$$\tau_A = I_A\frac{d\omega_A}{dt}$$
The gear B must turn at the same instantaneous tangential velocity. So their angular velocities are related. From here it is possible to get the relation between torques.
Because the force $f$ at the teeth (different from the applied force F) is the only source of torque at gear B:
$$f = \frac{\tau_B}{R_B}$$
But for gear A, there is F as the source of torque:
$$F - f = \frac{\tau_A}{R_A}$$
